# Polynomials: Number of Roots

1. Jun 26, 2010

### jgens

I'm trying to prove that a polynomial function of degree $n$ has at most $n$ roots. I was thinking that I could accomplish this by induction on the degree of the polynomial but I wanted to make sure that this would work first. If someone could let me know if this approach will work, I would appreciate it. Thanks!

2. Jun 26, 2010

### Office_Shredder

Staff Emeritus
Why don't you give it a shot and see what happens. Induction is probably the most common way to prove this

3. Jun 27, 2010

### mathman

One approach is going backwards. Given one root for a polynomial of degree n, you can use synthetic division to get a polynomial of degree n-1 for the rest of the roots. Continue until the polynomial is down to 0=0. You can do this exactly n times.

4. Jun 27, 2010

### HallsofIvy

Staff Emeritus
If $x_1$, $x_2$, ..., $x_n$ are the roots of a polynomial equation, then the polynomial is $a(x- x_1)(x- x_2)\cdot\cdot\cdot(x- x_n)$ for some number a. Multiplied out, that has degree n.

Notice that a polynomial of degree n has exactly n roots if we count "multiplicity" (if $x_1= x_2$ that still counts as "two roots") and if we count complex roots. Counting only distinct real roots, there still cannot be more than n roots.

5. Jun 27, 2010

### disregardthat

This is usually treated as a theorem of what OP's about to prove. I think it is at least most pedagogical to start out without the above, and rather let it be a corollary, thus keeping the 'natural order'.

6. Jun 28, 2010

### HallsofIvy

Staff Emeritus
What you could do, then, is prove that a polynomial of degree n has at least one (complex) solution. Then show that dividing the polynomial by x-a where a is that root, gives a polynomial of degree n-1. Since a polynomial cannot have degree less than 0, there can be, at most, n roots.

7. Jun 28, 2010

### Hurkyl

Staff Emeritus
An example of a ring where polynomials fail to have at most n roots is the ring of integers modulo 8. The polynomial f(x)=x2-1 has, in fact, 4 distinct roots: 1, 3, 5, and 7.

Proving there are at most n roots over a field is not a trivial thing. It is not a hard, but it is subtle -- easy to overlook relevant points.

Last edited: Jun 28, 2010
8. Jun 28, 2010

### disregardthat

Is not the fact that it is an integral domain sufficient?